# Growth and Decay of Current in an LR Circuit > i = I0(1 − e^(−tR/L)) with time constant L/R, 63.2 percent at one tau, inductor energy and switch-off sparking — JEE Physics transients. - Canonical URL: https://prepelephant.com/topics/jee/physics/lr-circuit-growth - Exam / course: JEE · Subject: Physics - Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team - First published: 2026-10-02 - Last updated: 2026-10-02 - How to cite: "Growth and Decay of Current in an LR Circuit", PrepElephant, https://prepelephant.com/topics/jee/physics/lr-circuit-growth ## Direct answer When a battery of emf V drives current through a resistor R and inductor L in series, the current cannot jump; it climbs as i = I0(1 − e^(−t/τ)) toward I0 = V/R with time constant τ = L/R, reaching 63.2 percent of its final value at t = τ. The inductor's voltage correspondingly decays as V e^(−t/τ), while the resistor's share rises to fill the gap. Open the circuit and the current decays exponentially, i = I0 e^(−t/τ), falling to 36.8 percent at one time constant. The energy ½LI² stored in the magnetic field is the reason for both behaviours: building it takes time the battery must pay for, and dumping it through an opened switch produces the sparking that JEE loves to explain. ## What you must remember - **Growth equation:** i = (V/R)(1 − e^(−Rt/L)); at t = τ = L/R the current is 63.2 percent of I0, at 2τ it is 86.5 percent, at 5τ it is within 1 percent — circuits have "settled" by five time constants by convention. - **Decay equation:** on removing the source (with a closed path), i = I0 e^(−t/τ); the half-life of the decay is t_half = τ ln 2 = 0.693 τ. - **Voltage split:** during growth V_L = V e^(−t/τ) and V_R = V(1 − e^(−t/τ)), so the inductor voltage can change discontinuously even though the current cannot. - **Time constant meaning:** τ = L/R is the time the current would take to reach I0 if it kept its initial rate di/dt = V/L; dimensionally henry/ohm is the second — a quick anchor for verifying any derived expression. - **Energy bookkeeping:** at any instant VI supplies power, i²R dissipates some, and the difference d(½Li²)/dt stores in the field; at t = τ with V = 20 V, R = 4 Ω the split is 63.2 W supplied, 40 W heated, 23 W stored. - **Switch-off physics:** interrupting an inductive circuit forces dI/dt toward infinity, so V = L dI/dt spikes across the break — the sparking, and the reason relay contacts arc and fluorescent starters exploit the surge. ## Following twenty volts through two henries Close a switch on V = 20 V, R = 4 Ω, L = 2 H. The final current is 5 A and the time constant τ = L/R = 0.5 s. At t = 0.5 s the current is 5(1 − e^(−1)) = 3.16 A, the resistor holds 12.6 V, and the inductor the remaining 7.4 V. Ask for the half-way current of 2.5 A and the logarithm appears: 2.5 = 5(1 − e^(−2t)) gives t = 0.347 s. The rate bookkeeping closes the picture at t = τ: the battery delivers Vi = 63.2 W, the resistor burns i²R = 40 W, and the missing 23.2 W is exactly d(½Li²)/dt = L i (di/dt) — every watt is audited. Now open the switch after steady state. The 2.5 J stored in the field (½ × 2 × 25) has nowhere to go but across the air gap, and because the current collapses in microseconds, the transient voltage V = L dI/dt reaches kilovolts — the visible spark. Practical circuits give the current a bypass (a flyback diode, or the bleeder path in fluorescent fittings) so the same energy dissipates gradually instead of an arc; the exam-ready statement is that the inductor opposes a change in current in both directions, violently at break and patiently at make. ## Where students slip JEE Main tests the anchors: current at one time constant (63.2 percent), the half-life relation t = τ ln 2, and identifying τ from a graph of i versus t. JEE Advanced pushes the bookkeeping: rates of energy storage at a stated instant, or a switch that moves the inductor between two resistive branches, changing τ mid-problem — the current at the switching instant is continuous, and that continuity is the only initial condition needed. The recurring errors: swapping L/R for RL; treating the inductor as an open circuit during the whole transient (open only at t = 0 for growth, a short at t = ∞); and reusing the old τ after the circuit's R has changed. One more classic: the inductor voltage jumps to the full emf at switching even though the current is zero — quantities that can jump (V_L) and quantities that cannot (i) partition the physics of every transient question. ## Frequently asked questions ### What is the current in an LR circuit one time constant after switching on? i = I0(1 − e^(−1)) ≈ 0.632 I0, so the current has reached 63.2 percent of its final value V/R. ### What does the time constant L/R physically represent? It is the time in which the current would reach its final value if it continued at its initial rate V/L — the natural clock of the exponential transient, in seconds. ### Why can't the current through an inductor change instantaneously? An abrupt current change would demand infinite V = L dI/dt, so the inductor enforces continuity of current even as its own voltage jumps discontinuously. ### What causes sparking when an inductive circuit is opened? The stored field energy ½LI² must leave through the break, forcing a huge dI/dt and a large voltage across the gap; a parallel diode or bleeder gives it a gentler path. ### Where does the missing battery power go during current growth? Into the magnetic field: at any instant VI minus i²R equals d(½Li²)/dt, and by steady state the battery has paid ½LI0² of stored energy plus all the resistive heat.